NCERT Solutions for Class 10 Maths Chapter 1 Ex 1.3: Exercise 1.3 of Chapter 1 Real Numbers in Class 10 Maths focuses on problems related to the division algorithm. This exercise helps students understand how to find the quotient and remainder when one number is divided by another, which is an important concept for working with real numbers.
The solutions provide detailed steps for applying the division algorithm to different pairs of numbers and help in understanding how to express the relationship between two numbers using division, quotient, and remainder. Solving these problems enhances students understanding of number theory and improves their ability to solve complex mathematical problems involving division. The solutions are a useful guide for exam preparation, ensuring a deeper comprehension of the topic.NCERT Solutions for Class 10 Maths Chapter 1 Ex 1.3 PDF
1. Prove that √ 5 is irrational.
Solutions: Let us assume that √ 5 is a rational number.
i.e., √ 5 = x/y (where, x and y are co-primes) y √ 5= x Squaring both sides, we get (y √ 5) 2 = x 2 ⇒5y 2 = x 2 ……………………………….. (1) Thus, x 2 is divisible by 5, so x is also divisible by 5. Let us say, x = 5k, for some value of k, and substituting the value of x in equation (1), we get 5y 2 = (5k) 2 ⇒y 2 = 5k 2 is divisible by 5, it means y is divisible by 5. Clearly, x and y are not co-primes. Thus, our assumption about √ 5 is rational is incorrect. Hence, √ 5 is an irrational number.2. Prove that 3 + 2√5 + is irrational.
Solutions: Let us assume 3 + 2 √ 5 is rational.
Then, we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y Rearranging, we get3. Prove that the following are irrational.
(i) 1/√2
(ii) 7√5
(iii) 6 + √ 2
Solutions:
(i) 1/ √ 2
Let us assume 1/√2 is rational. Then, we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y Rearranging, we get √2 = y/x x and y are integers; thus, √2 is a rational number, which contradicts the fact that √2 is irrational. Hence, we can conclude that 1/√2 is irrational.(ii) 7 √ 5
Let us assume 7√5 is a rational number. Then, we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y Rearranging, we get √5 = x/7y x and y are integers; thus, √5 is a rational number, which contradicts the fact that √5 is irrational. Hence, we can conclude that 7√5 is irrational.(iii) 6 + √ 2
Let us assume 6 +√2 is a rational number. Then, we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y Rearranging, we get √2 = (x/y) – x and y are integers; thus, (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number. Hence, we can conclude that 6 +√2 is irrational.